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Amanda [17]
2 years ago
10

Does the efficiency of a motor depend on mass?

Physics
2 answers:
Valentin [98]2 years ago
5 0

Practically yes

  • Efficiency=Output/input

So

If mass is more output may come less so it affects the efficiency practically

But thepritically it doesn't

fiasKO [112]2 years ago
5 0

Answer:

<u>Yes</u>

Explanation:

<u>Formula of motor efficiency</u>

  • η = Input Load (E₀)/ Output Load(E₁)

We know from simple physics that the energy of a body is directly proportional to its mass. Here, in the working of a motor, mass may not be involved directly, but the electrical energy required to provide a function will likely have more efficiency when working with lesser masses.

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A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-
solong [7]

Answer:

 N₁ = 393.96 N   and  N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

          N -W = m a

where the acceleration is ccentripeta

          a = v² / r

           

          N = W + m v² / r

          N = mg + mv² / r

         

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

           Em₀ = U = m g h

lowest point. Stop curl down

           Em_{f} = K = ½ m v²

           Emo = Em_{f}

           m g h = ½ m v²

           v² = 2 gh

we substitute

             N = m (g + 2gh / r)

            N = mg (1 + 2h / r)

let's calculate

          N₁ = 5 9.8 (1 + 2 17.6 / 5)

          N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

          -N -W = - m v₂² / r

            N = m v₂² / r - W

             N = m (v₂²/r  - g)

we seek speed with the conservation of energy

           Em₀ = U = m g h

final point. Top of circle with height 2r

             Em_{f} = K + U = ½ m v₂² + mg (2r)

              Em₀ =   Em_{f}

            mgh = ½ m v₂² + 2mgr

             v₂² = 2 g (h-2r)

we substitute

            N = m (2g (h-2r) / r - g)

            N = mg (2 (h-r) / r 1) = mg (2h/r  -2 -1)

             N = mg (2h/r  - 3)

            N = 5 9.8 (2 17.6 / 5 -3)

            N = 197.96 N

Directed down

3 0
3 years ago
A racing car is traveling on a circular race course of radius 569 m at a speed such that it makes 1.46 rev in 1 minute. the mass
HACTEHA [7]
1.46rev in 1min = 2pi*1.46revs/60s=0.152891 rad/s=ω
Fc=m*r*ω^2
Fc= 1550kg*(0.152891)^2 * 569m = 19951N 

hope this helps! Thanks!
8 0
2 years ago
This wave diagram represents a low-pitched, soft musical note.
garik1379 [7]

Answer:

A. lower peaks that are farther apart than those in the diagram

Explanation:

4 0
3 years ago
If a laptop of mass 1.4 kg is lifted to a table of height 0.8 m, how much gravitational potential energy is added to the laptop?
borishaifa [10]
Potential Energy = mgh , 

mass = 1.4 kg, g ≈ 10 m/s², h = 0.8m

Potential Energy = mgh = 1.4 * 10 * 0.8 = 11.2 N

Potential Energy added = 11.2 N

4 0
3 years ago
Read 2 more answers
A rocket test sled accelerates at a constant rate of 25 m/s2 from rest while moving along a straight track. The rocket engine tu
ddd [48]

Answer:

332.01 m

Explanation:

The total distance be the s = s_1+s_2+s_3

Case : 1 :- Acceleration

u = 0 m/s

v = 37 m/s

a = 25 m/s²

So, Applying equation of motion as:

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{37^2-0^2}{2\times 25}\\\Rightarrow s_1=27.38m

Case : 2 :- No Acceleration

t = 7 s

v = 37 m/s

s_2=v\times t\\\Rightarrow s=37\times 7\ m\\\Rightarrow s=259\ m

Case : 3 :- Deceleration

a = -15 m/s²

u = 37 m/s

v = 0 m/s

So, Applying equation of motion as:

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-37^2}{2\times -15}\\\Rightarrow s=45.63\ m

<u>Hence, S = 27.38 + 259 + 45.63 m = 332.01 m</u>

3 0
3 years ago
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